Question

How do you simplify (x+3) !
(x+2)!

Answer 1

They are dividing

Answer 2

is it $$\frac{(x+3)!}{(x+2)!}$$

Answer 3

yes

Answer 4

If so you can write this as,
$$\frac{(x+3)\times(x+2)!}{(x+2)!}$$
Since$$(x+3)!=(x+3)\times (x+2)!$$

Now I guess you can continue

Answer 5

I dont understand.

Answer 6

$$(x+3)!=(x+3)\times[(x+3)-1]\times[(x+3)-2]\times \cdots \times 1$$
I think you can understand this.
$$[(x+3)-1]=(x+2)\\ [(x+3]-2]=[(x+2)-1]\cdots$$ Thus
$$(x+3)!=(x+3)\times\underbrace{(x+2)\times[(x+2)-1]\times \cdots \times 1}_{(x+2)!}=(x+3)\times(x+2)!$$
Do you get it now?